For the purpose of getting fatter tails than the Guassian, I have seen people for example use $\alpha$-stable processes to model the stock. But in that case they end up using 'tempered' versions of the processes, where the tails decay exponentially so as to make the second moment finite. So the standard seems to be that the second moment must be finite. But why is this so? Is it just for tractability of the model or do they believe that finite second moment is an empirical fact?

Additional discussion:

In this paper Taleb explores the possibility of constructing a risk-neutral measure in an infinite-variance setting. As he mentions, this destroys the dynamic-hedging theory, but in practice it does not make a difference.

$\begingroup$I agree that pricing an option only requires to calculate its finite expectation under $Q$ as $E^Q(H/(1+r)^{T-t}|F_t)$ so the variance wouldnt matter. It may be that for hedging purposes you require a finite variance.$\endgroup$
– emcorJun 13 '15 at 13:42

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$\begingroup$I have heard it claimed that if stock returns have infinite variance, then a Call option (of any maturity) will have the same price as the stock. Which we don't see in real life. However, I am not sure if this statement is true or how to prove it.$\endgroup$
– Alex CJun 13 '15 at 20:46

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$\begingroup$@emcor It's certainly true that for hedging you need finite moments, or else the hedging error in discrete time doesn't necessarily converge to $0$ as $\Delta t \rightarrow 0$. But from what I can see, you could still use risk-neutral pricing just like one does in incomplete markets.$\endgroup$
– Slug PueJun 18 '15 at 13:20

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$\begingroup$Variance swaps trade at a finite price and not so high compared to option vols; in an infinite variance model the price would be infinite.$\endgroup$
– q.t.f.Jun 29 '15 at 20:44

$\begingroup$You do not see prices with big changes, going up or down, so the variance must be finite. Even if you see that, for tractability we need some kind of stationary series, so probably we wold model the second difference in prices.$\endgroup$
– RobertJun 30 '15 at 12:02

3 Answers
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I would argue that there is some path-dependency involved. The BS model is considered the big breakthrough and it presented the world with some kind of tractable toy model. After that people saw that you had to adjust the model to account for all kinds of stylized facts (e.g. non-constant volatility for different strikes, over time and so on). Yet finite variance somehow survived this tweaking of the models, may it be for mathematical convenience, may it be that people just didn't think into this direction because there are other, more important issues (like numerical stability and so on).

Another thing is that empirically variance is always finite and even theoretically stock prices can only fall to zero but not indefinitely so finite variance (even though as high as needed) seems like a not too bad idea.

Just another trivial reason is that most of statistics deals with probability distributions with finite variance and people tend to use what they find and what they are used to.

As an aside there is a new paper by Taleb where he develops an option pricing model which can cope with infinite variance:

Proof that under simple assumptions, such as constraints of Put-Call
Parity, the probability measure for the valuation of a European option
has the mean derived from the forward price which can, but does not
have to be the risk-neutral one, under any general probability
distribution, bypassing the Black-Scholes-Merton dynamic hedging
argument, and without the requirement of complete markets and other
strong assumptions. We confirm that the heuristics used by traders for
centuries are both more robust, more consistent, and more rigorous
than held in the economics literature. We also show that options can
be priced using infinite variance (finite mean) distributions.

On a pure technical aspect, a model does not need to have a finite variance. In the context of option pricing, what you need it a way to replicate the behaviour of the stock price. Once you have it you need to find a corresponding risk-neutral measure. There you will have the first difficulty, with infinite variance, the corresponding hedging strategy is not obious. However, as in the case of jumpe processes, you might still find a risk-neutral measure by means of min entropy for instance (see Fujiwara, Miyahara (2003)). So in theory, you can apply compute option prices using monte carlo.

On the other side, on a practical point of view, infinite variance distribution will cause many difficulties during the calibration of your model (simply because infinite variance are directly observable). Furthermore the standard stochastic calculs like Ito's-Lemma does not apply in this context. Infinte variance distribution model, that bring higher cost in term of complexity, should allow for better performences. However, the class of jump-diffusion processes (see Crepey (2013)) or even infinite levy process (see Tankov (2007)) are in pricipe rich enough to enable a efficient pricing (both in term of price replication as computation speed).

So, in my opinion, the reason why there are not used in practice is because there are not tractable enough the outperform more classical model.

I would also say that the pricing of some exotic products require to compute expectations of functions of the random variable at consideration, and these functions may grow more than linearly : you need finite moments in order for the prices of these exotic derivatives to be bounded.